Answer
Verified
36.9k+ views
Hint: In the above problem the trigonometric product to sum identities should be used. The identity to be used is to be determined by inspection so that the result gives angle with general known trigonometric ratio.
Given in the problem, we need to find value of the expression
$\sin {20^0}\sin {40^0}\sin {60^0}\sin {80^0}$ …………………………….. (1)
We need to group terms so that using a product to sum trigonometric formula gives angles whose trigonometric value is known.
We know that value of
$\cos (x - y) - \cos (x + y) = 2\sin x\sin y$ …………………………………...(2)
Put $x = 80$and $y = 40$in the equation (2) ,we get
\[
2\sin {80^0}\sin {40^0} = \cos ({80^0} - {40^0}) - \cos ({80^0} + {40^0}) \\
\Rightarrow 2\sin {80^0}\sin {40^0} = \cos ({40^0}) - \cos ({120^0}) \\
\]
We know that $\cos \left( {{{180}^0} - \theta } \right) = - \cos \theta $
Put $\theta = {60^0}$ in above $ \Rightarrow \cos {120^0} = \cos \left( {{{180}^0} - {{60}^0}} \right) = - \cos {60^0} = - \dfrac{1}{2}$
\[ \Rightarrow 2\sin {80^0}\sin {40^0} = \cos ({40^0}) - \cos ({120^0}) = \cos ({40^0}) + \dfrac{1}{2}\] …………………(3)
Multiplying expression (1) with $\dfrac{2}{2}$ and rearranging, we get
$\dfrac{{2\sin {{80}^0}\sin {{40}^0}\sin {{60}^0}\sin {{20}^0}}}{2}$
Using equation (3) in above, we get
$\left( {\cos {{40}^0} + \dfrac{1}{2}} \right)\dfrac{{\sin {{60}^0}\sin {{20}^0}}}{2}$
Using $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$ in above,
$ \Rightarrow \left( {\cos {{40}^0} + \dfrac{1}{2}} \right)\dfrac{{\sqrt 3 }}{4}\sin {20^0}$
$ \Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {2\sin {{20}^0}\cos {{40}^0} + \sin {{20}^0}} \right)$ …………………………………………………...(4)
We know that value of
$\sin (x + y) + \sin (x - y) = 2\sin x\cos y$ ……………………………………………...(5)
Put $x = 20$and $y = 40$in the equation (5), we get
\[
2\sin {20^0}\cos {40^0} = \sin ({20^0} + {40^0}) + \sin ({20^0} - {40^0}) \\
\Rightarrow 2\sin {20^0}\sin {40^0} = \sin ({60^0}) + \sin ( - {20^0}) \\
\]
We know that $\sin \left( { - \theta } \right) = - \sin \theta $
Put $\theta = {20^0}$ in above gives $\sin \left( { - {{20}^0}} \right) = - \sin {20^0}$
\[ \Rightarrow 2\sin {20^0}\sin {40^0} = \sin ({60^0}) - \sin ({20^0}) = \dfrac{{\sqrt 3 }}{2} - \sin ({20^0})\] ……………………………….(6)
Using equation (6) in (4), we get
\[
\dfrac{{\sqrt 3 }}{8}\left( {2\sin {{20}^0}\cos {{40}^0} + \sin {{20}^0}} \right) = \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2} - \sin {{20}^0} + \sin {{20}^0}} \right) \\
\Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{3}{{16}} \\
\]
Therefore, the value of expression (1) is $\dfrac{3}{{16}}$.
Hence option $(C)$ $\dfrac{3}{{16}}$ is the correct answer.
Note: Always remember trigonometric sum to product and product to sum formula. Modifications may need to be performed in the expressions like above in order to use these identities. These modifications should never alter the value of the original expression. Try to convert the expression in problems of above type into known trigonometric ratio values.
Given in the problem, we need to find value of the expression
$\sin {20^0}\sin {40^0}\sin {60^0}\sin {80^0}$ …………………………….. (1)
We need to group terms so that using a product to sum trigonometric formula gives angles whose trigonometric value is known.
We know that value of
$\cos (x - y) - \cos (x + y) = 2\sin x\sin y$ …………………………………...(2)
Put $x = 80$and $y = 40$in the equation (2) ,we get
\[
2\sin {80^0}\sin {40^0} = \cos ({80^0} - {40^0}) - \cos ({80^0} + {40^0}) \\
\Rightarrow 2\sin {80^0}\sin {40^0} = \cos ({40^0}) - \cos ({120^0}) \\
\]
We know that $\cos \left( {{{180}^0} - \theta } \right) = - \cos \theta $
Put $\theta = {60^0}$ in above $ \Rightarrow \cos {120^0} = \cos \left( {{{180}^0} - {{60}^0}} \right) = - \cos {60^0} = - \dfrac{1}{2}$
\[ \Rightarrow 2\sin {80^0}\sin {40^0} = \cos ({40^0}) - \cos ({120^0}) = \cos ({40^0}) + \dfrac{1}{2}\] …………………(3)
Multiplying expression (1) with $\dfrac{2}{2}$ and rearranging, we get
$\dfrac{{2\sin {{80}^0}\sin {{40}^0}\sin {{60}^0}\sin {{20}^0}}}{2}$
Using equation (3) in above, we get
$\left( {\cos {{40}^0} + \dfrac{1}{2}} \right)\dfrac{{\sin {{60}^0}\sin {{20}^0}}}{2}$
Using $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$ in above,
$ \Rightarrow \left( {\cos {{40}^0} + \dfrac{1}{2}} \right)\dfrac{{\sqrt 3 }}{4}\sin {20^0}$
$ \Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {2\sin {{20}^0}\cos {{40}^0} + \sin {{20}^0}} \right)$ …………………………………………………...(4)
We know that value of
$\sin (x + y) + \sin (x - y) = 2\sin x\cos y$ ……………………………………………...(5)
Put $x = 20$and $y = 40$in the equation (5), we get
\[
2\sin {20^0}\cos {40^0} = \sin ({20^0} + {40^0}) + \sin ({20^0} - {40^0}) \\
\Rightarrow 2\sin {20^0}\sin {40^0} = \sin ({60^0}) + \sin ( - {20^0}) \\
\]
We know that $\sin \left( { - \theta } \right) = - \sin \theta $
Put $\theta = {20^0}$ in above gives $\sin \left( { - {{20}^0}} \right) = - \sin {20^0}$
\[ \Rightarrow 2\sin {20^0}\sin {40^0} = \sin ({60^0}) - \sin ({20^0}) = \dfrac{{\sqrt 3 }}{2} - \sin ({20^0})\] ……………………………….(6)
Using equation (6) in (4), we get
\[
\dfrac{{\sqrt 3 }}{8}\left( {2\sin {{20}^0}\cos {{40}^0} + \sin {{20}^0}} \right) = \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2} - \sin {{20}^0} + \sin {{20}^0}} \right) \\
\Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{3}{{16}} \\
\]
Therefore, the value of expression (1) is $\dfrac{3}{{16}}$.
Hence option $(C)$ $\dfrac{3}{{16}}$ is the correct answer.
Note: Always remember trigonometric sum to product and product to sum formula. Modifications may need to be performed in the expressions like above in order to use these identities. These modifications should never alter the value of the original expression. Try to convert the expression in problems of above type into known trigonometric ratio values.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
Other Pages
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main
The cell in the circuit shown in the figure is ideal class 12 physics JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main
An electric bulb has a power of 500W Express it in class 11 physics JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main